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Question
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
Solution
Let I = `int_0^(pi/4) log(1 + tanx) "d"x` ......(i)
= `int_0^(pi/4) log[1 + tan(pi/4 - x)] "d"x` ......`[∵ int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x) "d"x]`
= `int_0^(pi/4) log(1 + (tan pi/4 - tan x)/(1 + tan pi/4 * tanx)) "d"x `
= `int_0^(pi/4) log (1 + (1 - tan x)/(1 + tan x)) "d"x`
= `int_0^(pi/4) log ((1 + tanx + 1 - tan x)/(1 + tan x)) "d"x`
= `int_0^(pi/4) log(2/(1 + tan x)) "d"x`
= `int_0^(pi/4)[log2 - log(1 + tanx)] "d"x`
= `log 2 int_0^(pi/4) 1* "d"x - int_0^(pi/4) log(1 + tanx) "d"x`
∴ I = `log 2[x]_0^(pi/4) - "I"` ......[From (i)]
∴ 2I = `log 2(pi/4 - 0)`
∴ I = `pi/8 log 2`
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